CONTINUITY AND DIFFERENTIABILITY

Transcription

1 CONTINUITY AND DIFFERENTIABILITY Revision Assignment Class 1 Chapter 5 QUESTION1: Check the continuity of the function f given by f () = 7 + 5at = 1. The function is efine at the given point = 1 an its value is 1. Let fin the limit of the function at = 1. lim f = lim = =1 1 1 lim f = f (1) 1 Hence, f is continuous at = 1. QUESTION: For what value of k is the following function continuous at = 1 1, 1 f = 1 4 k, = 1 1, 1. f = 1 4 k, = 1 1 ( 1)( + 1) lim f = lim = lim = lim( + 1) = f (1) = 4 k f() is continous at = 1 lim f = f (1) 1 = 4k k = 1 QUESTION:Show that a cosine function is a continuous function

2 f = cos lim cos = lim cos( a + h) a h 0 h 0 ( a a ) = lim cos cosh sin sinh = lim cos a cosh lim sin a sinh h 0 h 0 = cos a lim cosh sin a lim sinh h 0 h 0 = cos a 1 sin a 0 = cos a. f ( a) = cos a lim f = f ( a) a QUESTION 4: Fin two points at which the given function y = is continous but not ifferentiable., 0 A moulus function y = = is continous but not ifferentiable at = 0 -,<0 -, - = is continous but not ifferentiable at = -( - ), < -5, 5 an -5 = is continous but not ifferentiable at = 5 -( -5 ), <5 Consier a cobination of two such functions y = This function will be continous but not ifferentiable at = an at = 5 QUESTION 5: Use Mathematical Inuction to prove that. n n n 1 =, where n is a natural number

3 n n 1 LetP( n) : ( ) = n P ( ) = = = P(1) is true (1) : Let us assume P(k) k k 1 P( k) : ( ) = k Using P(k), we shall prove P(k+1) LHS = = k k = ( ) + ( ) k 1 k. ( k ) ( ) k k ( k ) ( ) k + 1 ( k ) k P( k + 1) : = ( k + 1) = n. = + = + k = ( k + 1) = RHS QUESTION : k + 1 k (. ).1 8 ( ) 7 = 8( ) 7 8( ) (18 9) = + Differentiate( ) ( ) 7 8.(9 9 5) ( 1) = + 7 1(9 9 5) ( 1) 8 QUESTION 7: D 9 ifferentiate(sin + cos )

4 9 (sin + cos ) 9 sin cos = + 8 sin 5 cos = 9sin + cos = sin (cos ) cos ( sin ) 8 5 = 9sin cos cos sin = 7 4 sin cos (sin cos ) QUESTION 8: Fin [ cos( cos + sin ) ] [ cos( cos + sin ) ] = sin( cos + sin ) ( cos + sin ) = sin( cos + sin )(sin + cos ) QUESTION 9: Using the fact that sin (A - B) = sin A cos B - cos A sin B an the ifferentiation, obtain the ifference formula for cosines. We shall take A to be a constant an ifferentiate w.r.t sin(a-b) = sin A cos B - cos A sin B [ sin(a-b) ] = [ sin A cos B - cos A sin B] cos( A B) (-B) = sin A cos B - cos A sin B cos( A B) B = sin A (-sinb) B - cos A cosb B cos( A B) B = [ -cos A cosb - sin A sinb ] B [ ] [ ] cos( A B) = - cos A cosb + sin A sinb cos( A B) = cos A cosb + sin A sinb

5 QUESTION 10:. Lety = (7 ) cos log y = cos log(7 ) Differentiate(7 ) [ cos log(7 )] cos log y 1 7cos log(7 ).( sin ) y cos log(7 )sin y cos y log(7 )sin cos (5 ) cos log(7)sin [ ] 1 QUESTION 11: Differentiate sin ; 0 1

6 1 sin 1 = 1 ( ) = = = = = 1 1 cos QUESTION 1: Differentiate 4 + 7

7 1 cos = cos cos = ( ) cos = = cos cos = cos Fin if y = sin sin 1 when0 < < QUESTION 1:

8 Now = - or + ( ) ( 1 ) 1 1 = ( ) ; when0 < < ; when 1< < ; when 1< < 0 ; when0 < < 1 1 when0 < < 1, = 1 Fin, if y = log5, > 1 QUESTION 14: log5 log5 y = log5, > 1 log5 log y = log log5 log y = log 5 log(log 5 ) 1 y = { log 5 log ( log5) } ( log5 )..5 log(log 5 ).5 y = + log log(log 5 ) y = + y { 1+ log(log 5 ) } { } log5 ( log5) { 1+ log(log 5 ) } sin + cos QUESTION 15: Fin the erivative of y = (sin + cos ) ;0 < < π

9 Let y = (sin + cos ) sin + cos sin + cos log y = log (sin + cos ) log y = (sin + cos ) log(sin + cos ) 1 (sin + cos ) log(sin cos ) log(sin cos ) (sin cos ) y (sin + cos ) (sin y + cos ) + log(sin + cos ) (sin + cos ) (sin + cos ) 1 1 (sin + cos ) (cos sin ) log(sin cos ) (cos sin ) y (sin cos ) (cos sin ) + (cos sin ) log(sin + cos ) y 1 (cos sin ) 1+ log(sin + cos ) y [ ] y(cos sin ) [ 1+ log(sin + cos ) ] sin cos (sin cos ) + + (cos sin ) 1+ log(sin + cos ) QUESTION 1: 4 4 Fin. { } at = [ ] [ ]

10 { } = : Let u = log u = log 1 u log + log u 1 u 1 + log.1 u 1 u 1+ log u u [ 1+ log ] 4 = 4 4 = 4 log = [ 1 log ] 4 = [ 1+ log ] log = = 9[ 1+ log ] log 4 = 9[ 1+ log ] + 4[7 + 1log 4] = QUESTION 17: Fin. + ( ) for = 5 4 log 4

11 + ( ) = + ( ) = u + v( say) where, = u;;( ) = v u = : log u = log 1 u + u 1 u 1 ( ) + log u u 1 ( ) log + Let v = ( ) ( ) log log ( ) log v log( ) 1 v log( ) + log( ) v 1 v 1 + log( ) v v ( ) + log( ) 1 ( ) + = ( ) + log + ( ) + log( ) At =5: ( + ) = 5 ( 5 ) + (.5) log 5 + (5 ) + (.5) log(5 ) = 5 ( 5 ) + (.5) log 5 + (5 ) + (.5) log(5 ) = log log() 5 QUESTION 18:

12 Fin, if cos y = cos(0 + y) cos y = cos(0 + y) cos y = cos(0 + y) cos(0 + y) cos y cos y cos 0 + y cos ( 0 + y) + y y y ( + y) cos ( 0 + y) y ( + y) cos ( 0 + y) + y y ( + y y) ( + y) ( 0 + y) cos ( 0 + y) sin ( 0) sin 0 cos ( 0 + y) cos(0 )( sin ) cos ( sin 0 ) cos sin 0 cos(0 )(sin ) sin 0 cos 0 sin 0 cos 1 1 = = QUESTION 19: π π Fin if y = 10(1 cos t); = 15( t sin t); < t <

13 y = ; Fin π π y = 10(1 cos t); = 15( t sin t); < t < = 10(0 + sin t); = 15(1 cos t) t t t t 10(sin cos ) t 10(sin t) t = = = = cot 15(1 cos t) t 15(sin ) t

14 ( ) 0 + = ( 0 ) + = ( 0) + = 0 ( 0) ( 0) y = y y y 0 0 ( ).1 0 = ( 0) ( 0) = = { y + } { } ( ) + 0 But 0 + = y ( 0) y 5) Now, = y = ( 5) ( 0) = = QUESTION 0:

15 y Fin ; = 0(cost + t sin t) an y = 0 (sint - t cost) 0( sin t + t cost + sin t) = 0( t cos t) t an = 0 (cost - t (-sint ) - cos t)=0 ( t sint ) t t 0 ( t sint ) sin t = = = tan t 0( t cos t) cost t y t sec t sec t = (tan t) = = = 0tcos t 0t QUESTION 1: '' Fin f for f() = 7 such that R an < 0, 0 We know, =, < 0 7, 0 f = 7 = < / 1, 0 f = < // f = < 7, 0 1, 0 4, 0 4, 0 '' f for f() = an < is